1. Thermodymanics Lecture 3 8/31/2004

2. Units

3. Energy Ultimate source of energy is the sun E = h 57 Kcal/mol of photons green light or 238 KJ/mol 1 cal = 4.184 joules 0.239 cal = 1 J ATP energy carrier, for hydrolysis to ADP + Pi = 7.3 kcal/mole or 30.5 KJ/mol While vibrational energy infrared) = 0.6 kcal/mol or 2.5 KJ/mol C - C bond = 83 Kcal/mol or 346 KJ/mol the framework of a carbon skeleton is thermally stable but non-covalent bonds are only a few kcal/mol or 10-20 KJ/mole

4. Thermal Noncovalent ATP Green C-C glucose bond light bond 1 10 100 1000 Kcal/mol KJ/mol 1 10 100 1000 Biomolecule shapes and interactions are mediated by 4 types of non-covalent bonds. These bonds are responsible for the overall shape and interaction among biomolecules and can be modified by thermal energy.

5. Thermodynamics: Allows a prediction as to the spontaneous nature of a chemical reaction: Will this reaction proceed in a forward direction as the reaction is written: A+BC Will A react with B to form C or not? Is this reaction going from higher to lower energy? ENERGY not necessarily just heat!!!

6. Definitions: System: a defined part of the universe a chemical reaction a bacteria a reaction vessel a metabolic pathway Surroundings: the rest of the universe Open system: allows exchange of energy and matter Closed system: no exchange of matter or energy. i.e. A perfect insulated box.

7. Direction of heat flow by definition is most important. q = heat absorbed by the system from surroundings If q is positive reaction is endothermic system absorbs heat from surroundings If q is negative exothermic system gives off heat. A negative W is work done by the system on the surroundings i.e expansion of a gas

8. First Law of Thermodynamics: Energy is Conserved Energy is neither created or destroyed. In a chemical reaction, all the energy must be accounted for. Equivalence of work w and energy (heat) q Work (w) is defined as w = F x D (organized motion) Heat (q) is a reflection of random molecular motions (heat)

9.  U = U final - U initial = q - w Exothermic system releases heat = -q Endothermic system gains heat = +q  U = a state function dependent on the current properties only.  U is path independent while q and w are not state functions because they can be converted from one form of energy to the other. (excluding other forms of energy, e.g. electrical, light and nuclear energy, from this discussion.)

10. Enthalpy (H) At constant pressure w = P  V + w 1 w 1 = work from all means other than pressure-volume work. P  V is also a state function. By removing* this type of energy from U, we get enthalpy or ‘to warm in’ *remember the signs and direction H = U + PV  H =  U + P  V = q p -w + P  V = q p - w 1

11. Enthalpy (H) When considering only pressure/volume work  H = q p - P  V + P  V = q p  H = q p when other work is 0 q p is heat transferred at constant pressure. In biological systems the differences between  U and  H are negligible (e.g. volume changes)

12. The change in enthalpy in any hypothetical reaction pathway can be determined from the enthalpy change in any other reaction pathway between the same products and reactants. This is a calorie (joule) “bean counting” HotCold Which way does heat travel? This directionality, is not mentioned in First Law

13. Entropy (S) is the arrow of time The entropy of the universe is always increasing. A change in Entropy does not change the total energy. What does this mean in a closed system? What does this mean in an open system?

14. Disorder increases N identical molecules in a bulb, open the stop cock you get 2 N equally probable ways that N molecules can be distributed in the bulb. Gas on its own will expand to the available volume.

15. As the gas molecules are indistinguishable only N+1 different states exist in the bulb or 0, 1, 2, 3, 4,…., (N-1) molecules in left bulb W L = number of probable ways of placing L of the N molecules in the left bulb is The probability of occurrence of such a state (Is its fraction of the total number of states)

16. Direction occurs because the aggregate probability of all other states is only the most probable state will occur. for N = 10 23 Possible arrangements

17. Each molecule has an inherent amount of energy which drives it to the most probable state or maximum disorder. k b or Boltzman’s constant equates the arrangement probability to calories (joules) per mole. Entropy is a state function and as such its value depends only on parameters that describe a state of matter. Entropy (S) measure the degree of randomness

18. The process of diffusion of a gas from the left bulb initially W 2 = 1 and S = 0 to Right (N/2) + Left (N/2) at equilibrium gives a :  S that is (+) with a constant energy process such that  U = 0 while  S>0 This means if no energy flows into the bulbs from the outside expansion will cool the gas! Conservation of Energy says that the increase in Entropy is the same as the decrease in thermal (kinetic) energy of the molecules!!

19. It is difficult or (impossible) to count the number of arrangements or the most probable state! So how do we measure entropy? It takes 80 kcal/mol of heat to change ice at zero °C to water at zero °C 80,000 = 293 ev’s or entropy units 273 A Reversible process means at equilibrium during the change. This is impossible but makes the calculations easier but for irreversible process

20. At constant pressure we have changes in q p (Enthalpy) and changes in order - disorder (Entropy) A spontaneous process gives up energy and becomes more disordered. G = H - TS Describes the total usable energy of a system, thus a change from one state to another produces:  G =  H - T  S = qp - T  S If  G is negative, the process is spontaneous

21.  S  H + -All favorable at all temperature spontaneous - -Enthalpy favored. Spontaneous at temperature below T =  H  S + +Entropy driven, enthalpy opposed. Spontaneous at Temperatures above T =  H  S -+Non-spontaneous

22. STP Standard Temperature and Pressure and at 1M concentration. We calculate  G’s under these conditions. aA + bB cC + dD We can calculate a G for each component (1) (2) combining (1) and (2) (3)

23. Now if we are at equilibrium or  G = 0 Then So what does  G o really mean? OR

24. K eq GoGo G If K eq = 1 then  G = 0 G o equates to how far K eq varies from 1!!

25. K eq can vary from 10 6 to 10 -6 or more!!!  G o is a method to calculate two reactions whose K eq ’s are different However The initial products and reactants maybe far from their equilibrium concentrations so Must be used

26. Le Chatelier’s principle Any deviation from equilibrium stimulates a process which restores equilibrium. All closed systems must therefore reach equilibrium What does this mean? Keq varies with 1/T If K eq varied with temperature things would be very unstable. Exothermic reactions would heat up causing an increase in K eq generating more heat etc….

27. Fortunately, this does not happen in nature. R = gas constant for a 1M solution Plot lnKeq vs. 1/T ( remember T is in absolute degrees Kelvin) lnKeq = Slope = Intercept Van’t Hoff plot

28. K eq  G o (kJ·mole-1 10 6 -34.3 10 4 -22.8 10 2 -11.4 10 1 -5.7 10 0 0.0 10 -1 5.7 10 -2 11.4 10 -4 22.8 10 -6 34.3 The Variation of K eq with  G o at 25 o C

29. The f = for formation. By convention, the free energy of the elements is taken as zero at 25 o C The free energies of any compound can be measured as a sum of components from the free energies of formation.

30. Standard State for Biochemistry Unit Activity 25 o C pH = 7.0 (not 0, as used in chemistry) [H 2 O] is taken as 1, however, if water is in the Keq equation then [H 2 O] = 55.5 The prime indicates Biochemical standard state

31. Most times However, species with either H 2 O or H + requires consideration. For A + B C + D + nH 2 O This is because water is at unity. Water is 55.5 M and for 1 mol of H 2 O formed:

32. For: A + B C + HD H + + D - Where a proton is in the equation K This is only valid for

33. Coupled Reactions A + B C + D  G 1 (1) D + E F +G  G 2 (2) If reaction 1 will not occur as written. However, if is sufficiently exergonic so Then the combined reactions will be favorable through the common intermediate D A + B + E C + F + G  G 3

34. As long as the overall pathway is exergonic, it will operate in a forward manner. Thus, the free energy of ATP hydrolysis, a highly exergonic reaction, is harnessed to drive many otherwise endergonic biological processes to completion!!